Find the center, vertices,
foci, eccentricity, and asymptotes of the hyperbola with the given equation,
and sketch:

Since the y part
of the equation is added, then the center, foci, and vertices will be
above and below the center (on a line paralleling the y-axis),
rather than side by side.

Looking at the denominators, I see that
a2 =
25 and b2
= 144, so a
= 5 and b
= 12. The equation c2
– a2 = b2
tells me that c2
= 144 + 25 = 169, so c
= 13, and the eccentricity is e
= 13/5. Since x2
= (x – 0)2 and
y2 =
(y – 0)2, then
the center is at (h,
k) = (0, 0). The vertices
and foci are above and below the center, so the foci are at (0,
–13) and (0,
13), and the vertices are at (0,
5) and (0,
–5).

Because the y
part of the equation is dominant (being added, not subtracted),
then the slope of the asymptotes has the a
on top, so the slopes will be m
= ± 5/12. To graph, I start
with the center, and draw the asymptotes through it, using dashed
lines:

Then I draw in the vertices,
and rough in the graph, rotating the paper as necessary and "eye-balling"
for smoothness:

Since the a2
went with the x part
of the equation, then a
is in the denominator of the slopes of the asymptotes, giving me m
= ± 3/4. Keeping in mind that the
asymptotes go through the center of the hyperbola, the asymptes are
then given by the straight-line equationsy – 2 = ± (3/4)(x
+ 3).

Then the center is at (h,
k) = (–5, –3). Since the x
part of the equation is added, then the center, foci, and vertices lie
on a horizontal line paralleling the x-axis;
a2 =
25 and b2
= 20, so a =
5 and b
= 2sqrt[5]. The equation a2
+ b2 = c2
gives me c2
= 25 + 20 = 45, so c
= sqrt[45] = 3sqrt[5].
The slopes of the two asymptotes will be m
= ± (2/5)sqrt[5]. Then my
complete answer is:

center (–5,
–3), vertices
(–10,
–3) and (0,
–3),

foci
and ,

and asymptotes

If I had needed to graph this hyperbola,
I'd have used a decimal approximation of ±
0.89442719... for the slope, but would
have rounded the value to something reasonable like m
= ± 0.9.